2. Inequalities Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Inequalities T1 Inequalities 1 2 4 132 x xx   ----------- (*) 2 4 1032 x xx   2 22 4 3 2 03 2 3 2 x x x x x x x       2 2 1 032 xx xx   2 2 1 023 xx xx   Since 2 2 1310 24x x x       for any real x (OR coefficient of 2 10x  and discriminant, 22 4 ( 1) 4(1)(1) 3 0b ac      imply that 2 10xx   for any real x) 2 1 023xx  1 0( 1)( 3)xx  1 or 3xx   By replacing x with   2x in (*), we have: 2 24 4 132 x xx   2 42 4 123 x xx   which is what we need to solve. From earlier part, 22 1 or 3xx    2 1x  1 or 1xx  – 1 3 x + + – (N.A., since 2 0x for all real x)

River Valley High School, Mathematics Department 2023 Inequalities T2 2i   224 4 3 2 1 2 0x x x x        Alternative Method Discriminant =    24 4 4 3 32  Since coefficient of 2x is positive and discriminant < 0, 24 4 3xx is positive ii Therefore or iii Let lnyx 32 2 4 4 3 02 y y y yy   20 y   or 1y 2 ln 0 x   or ln 1x  1x or ln 1x or ln 1x 11 or 0 e or ex x x     3 2 25 03 w w   OR 33 w   or 5 2w   2 2 5 sin 03 yx y   , Since sin 0x for 3 2x   , 2 25 03 y y   From above, 33 y   or 5 2y  03 y or 5 2y            32 2 2 2 4 4 3 02 4 4 3 021 0 since 2 1 2 021 x x x xx x x x xx x xxxx           2x 01 x 0 1 + − + -2 − – + – + 3 3 5 2 3 3 5 2

River Valley High School, Mathematics Department 2023 Inequalities T3 33 y   or 5 2y or 5 2y 4i         2 2 44 2 4 402 4 ( 2) 4 02 24 02 ( 1) 5 0( 2) 11 02 x x xx xx x xx x x x xx x                     2 or 1 1xx        Alternative       2 2 2 2 44 2 Multiply by ( 2) on both sides, (4 )( 2) 4( 2) 0, 2 ( 2)[(4 )( 2) 4] 0 ( 2)( 2 4) 0 ( 2)( 2 4) 0 ( 2) 1 1 0 x x x x x x x x x x x x x x x x x x x                                2 or 1 1xx        ii 44 2x x  – + – + 22 ( )( )a b a b a b    2x

River Valley High School, Mathematics Department 2023 Inequalities T4       441 12 45 1 Hence, replace with 1 in earlier sol: 1 2 or 1 1 1 1 or 2 2 as 0 2 and 2 for all real as 0 and (2 ) x x x x xx xx xx x x x x x x                                       No solution 2 i.e. 2 2 x x          5   2